我想找出我的数据的每一列中NaN的数量。
当前回答
你可以试试:
In [1]: s = pd.DataFrame('a'=[1,2,5, np.nan, np.nan,3],'b'=[1,3, np.nan, np.nan,3,np.nan])
In [4]: s.isna().sum()
Out[4]: out = {'a'=2, 'b'=3} # the number of NaN values for each column
如果需要nan的总和:
In [5]: s.isna().sum().sum()
Out[6]: out = 5 #the inline sum of Out[4]
其他回答
希望这能有所帮助,
import pandas as pd
import numpy as np
df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan],'c':[np.nan,2,np.nan], 'd':[np.nan,np.nan,np.nan]})
df.isnull().sum()/len(df) * 100
Thres = 40
(df.isnull().sum()/len(df) * 100 ) < Thres
如果只是在pandas列中计算nan值,这里是一个快速的方法
import pandas as pd
## df1 as an example data frame
## col1 name of column for which you want to calculate the nan values
sum(pd.isnull(df1['col1']))
我写了一个简短的函数(Python 3)来生成.info作为pandas数据框架,然后可以写入excel:
df1 = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan]})
def info_as_df (df):
null_counts = df.isna().sum()
info_df = pd.DataFrame(list(zip(null_counts.index,null_counts.values))\
, columns = ['Column', 'Nulls_Count'])
data_types = df.dtypes
info_df['Dtype'] = data_types.values
return info_df
print(df1.info())
print(info_as_df(df1))
这使:
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 3 entries, 0 to 2
Data columns (total 2 columns):
# Column Non-Null Count Dtype
--- ------ -------------- -----
0 a 2 non-null float64
1 b 1 non-null float64
dtypes: float64(2)
memory usage: 176.0 bytes
None
Column Nulls_Count Dtype
0 a 1 float64
1 b 2 float64
使用isna()方法(或者它的别名isnull(),这也兼容较旧的pandas版本< 0.21.0),然后求和来计算NaN值。其中一列:
>>> s = pd.Series([1,2,3, np.nan, np.nan])
>>> s.isna().sum() # or s.isnull().sum() for older pandas versions
2
对于一些列,这也适用:
>>> df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan]})
>>> df.isna().sum()
a 1
b 2
dtype: int64
import pandas as pd
import numpy as np
# example DataFrame
df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan]})
# count the NaNs in a column
num_nan_a = df.loc[ (pd.isna(df['a'])) , 'a' ].shape[0]
num_nan_b = df.loc[ (pd.isna(df['b'])) , 'b' ].shape[0]
# summarize the num_nan_b
print(df)
print(' ')
print(f"There are {num_nan_a} NaNs in column a")
print(f"There are {num_nan_b} NaNs in column b")
给出输出:
a b
0 1.0 NaN
1 2.0 1.0
2 NaN NaN
There are 1 NaNs in column a
There are 2 NaNs in column b
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